Multiplicating inequalities I have two inequalities: $|x|\leq\sqrt{x^2+y^2}$ and $|y|\leq\sqrt{x^2+y^2}, \forall x,y \in \Bbb R$, can I multiply these inequalities to get $|xy|\leq x^2+y^2$?
If yes, what is the justification? If not, why?
 A: You can multiply these inequalities, because both sides of both inequalities are nonegative. Here is an explaination:
$$|x| \leq \sqrt{x^2+y^2}$$
$|y| \geq 0$, so you can multiply both sides by $|y|$:
$$|xy| \leq |y|\sqrt{x^2+y^2}$$
But we know that $|y| \leq \sqrt{x^2+y^2}$ and $0 \leq \sqrt{x^2+y^2}$, so:
$$|y|\sqrt{x^2+y^2} \leq \sqrt{x^2+y^2} \times \sqrt{x^2+y^2}=x^2+y^2$$
Finally:
$$|xy| \leq |y|\sqrt{x^2+y^2} \leq x^2+y^2$$
A: The answer is yes, because everything involved ($|x|$, $\sqrt{x^2+y^2}$, $|y|$), is positive. 
For a formal proof, begin by letting $P$ be the set of positive numbers, which is closed under addition and multiplication (i.e. if $a$ and $b$ are in $P$, so are $a+b$ and $ab$), and let $a<b$ mean $b-a\in P$. We will show that 
$$0\leq a<b\quad\text{and}\quad 0\leq c<d \implies ac<bd.$$
Firt, since $a<b$, $b-a$ is in $P$ and since $c$ is also in $P$, $c(b-a)$ is in $P$. Similarly, $b(d-c)$ is in $P$. Summing gives $c(b-a)+b(d-c)=bc-ac+bd-bc=bd-ac$ is in $P$, so we must have $bd>ac\implies ac<bd$ as desired.
A: Formally, what you use is that if you multiply an (non-strict) inequality by a nonnegative number, the inequality is preserved. 
So, if $0\leq a\leq b$ and $0\leq c\leq d$, then $ac\leq bc $ (multiplying the first inequality by $c$). If you now multiply the second inequality by $b$, you get $bc\leq bd$. Thus
$$
ac\leq bc\leq bd,
$$
and then $ac\leq bd$. 
A: The answer is a strong YES , because if xy is negative ,the inequality xy<=x^2+y^2 is trivial . And if xy is positive then our inequality is equivalent to (x-y/2)^2+(3/4)y^2 which is evidently positive 
